Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 3 · Variant 2

1 (a) Sketch the graph of y = x + 3a , where a is a positive constant. [1] (b) Hence or otherwise solve the inequality x + 3a 2 a - 2 x . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) y B1 Roughly symmetrical. Condone some inaccuracy, but needs to look as if they intended symmetry and straight lines. If not, then B0. Needs to be in the correct position. 3a Needs to have two solid line segments. Needs to exist in both quadrants above the axis. Ignore dotted lines below the axis. Solid line below the axis is B0. -3a O x Condone if no scale shown, but need to see 3a and -3a marked. Ignore y = a − 2 x if seen. 1 1(b) Obtain critical value − 23a from x + 3a = a − 2x B1 Ignore x = 4a if seen. State final answer x − 23a B1 Need a clear conclusion – must imply rejection of x = 4a. B0 if using ⩾. Alternative Method for Question 1(b) Obtain critical value − 23a from ( x + 3a ) 2 = ( a − 2 x ) 2 B1 Ignore x = 4a if seen. State final answer x − 23a B1 Need a clear conclusion – must imply rejection of x = 4a. B0 if using ⩾. 2

More questions on Algebra

2 Solve the equation 3 # 2 x + 1 = 4 # 3 2 x - 3 . Give your answer correct to 3 significant figures. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Use a correct law for the logarithm of a product or the logarithm of a power *M1 Not available after incorrect manipulation, e.g. 6 x +1 = 12 2 x − 3 Obtain ln3 + ( x + 1) ln 2 = ln 4 + ( 2 x − 3 ) ln3 A1 Or equivalent with no powers Solve for x DM1 ln 812 E.g. x = ln 92 Allow with logarithms evaluated. Obtain 2.46 only A1 Alternative Method for Question 2 Use a correct law of indices for a product or a power *M1 x +1 x E.g. 2 = 2  2 Not available after incorrect manipulation, e.g. 6 x +1 = 12 2 x − 3. 4 = 2 2 on its own is not enough. 2 4  9 x = 2 A1 OE with powers of x only, e.g. 6  2 x = 27 ) x Obtain ( 9 81 2 Note: 3x is not far enough. ( ) Solve for x DM1 ln 812 2 or x = ln E.g. x = 2 81 . 2 9 ln 9 Allow with logarithms evaluated. Obtain 2.46 only A1 4

More questions on Logarithmic and exponential functions

3 Im i O 2 3 Re P The shaded region in the Argand diagram, bounded by a line and a circle, represents the complex numbers z satisfying Rez G 2 and z - ( 3 + )i G 2 . The point P shown on the diagram is one of the points of intersection of the line and the circle. (a) Find the complex number represented by the point P. Give your answer in the form x + iy , where x and y are real and exact. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the greatest value of argz for points in the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Complete method to obtain equation in y only M1 E.g.( 2 − 3) 2 + ( y − 1) 2 = 4 or Obtain the y coordinate of P Or consider triangle and state  3 − 1 . ( ) 1 3 2 A1 Or exact equivalent in the form x + iy. Obtain 2 + i 1 − 3 ( ) Allow 2 − i 3 − 1 . ( ) Must not be coordinates, and not x, y stated separately. 2 3(b) −1 1 −1 2 B1 Im(z) State one relevant angle, e.g. tan or sin 3 10 2 i 10 1 O 3 Re(z) −1 1 −1 1 Note: tan = sin 3 10 Complete method to obtain the required angle M1 −1 1 −1 2 tan + sin 3 10 Obtain 1.01 radians or 57.7 A1 Accept AWRT Alternative Method for Question 3(b) 2 2 B1 Must be choosing the positive root. If y = mxis a tangent to the circle  ( x − 3) + ( mx − 1) = 4  1 + m 2 , ) ( ( x 2 + ( −−6 2 m ) x + 6 = 0 ) 2 2 3 + 2 6 1 + m = 0 and m = then ( −−6 2 m ) − 24 ( ) 5 Substitute into the quadratic and solve for x, or use gradient to obtain tangent M1 9 − 6 Note x =  9 − 6  5 π −1  5  Required angle is − sin OE  2   6    Obtain 1.01 radians or 57.7 A1 Accept AWRT 3

More questions on Complex numbers

1 4 Find the exact value of x tan -1 x dx . [6] y 0 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 2 −1 x 2 M1* Condone sign error in formula for integration by parts. Begin integration by parts and obtain px tan x + q dx 2  1 + x x 2 −1 1 x 2 A1 OE Obtain tan x − dx 2 Allow with arctanx or tan-1 x. 2 2  1 + x DM1 Split the fraction and integrate. x 2  Use dx and integrate to obtain x + tan −1 x  2 dx =  1 − 1 2  1 + x  1 + x A1 OE x 2 −1 1 −1 Obtain tan x − x − tan x ( ) 2 2 Substitute correct limits correctly in an expression of the form DM1 π 1 π 2 −1 −1 − 0 − + + 0 ( −0 ) x tan x + x + tan x 8 2 8 Dependent on both previous M marks. Need to see evidence of the use of the lower limit at least once. Must evaluate the trigonometry using radians. π 1 A1 ISW Obtain − Or exact two term equivalent 4 2 6

More questions on Integration

5 (a) It is given that f ( x) = ( x - a ) 2 g ( x) , where f ( x) and g ( x) are polynomials. Show that ( x - a ) is a factor of fl( )x . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) It is given that ( x - 3 ) 2 is a factor of 2x 3 - 4 x 2 + px + q , where p and q are constants. Find the values of p and q. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) Use correct product rule M1 2 f  ( x ) = ( x − a ) g ( x ) + 2 ( x − a ) g ( x ) Allow incorrect chain rule. Obtain correct derivative and state a clear conclusion A1 AG E.g. take out a factor of (x – a) and show the factorised form as far as ( x − a )h( x ) correctly (no further comment needed), or show that f  ( a ) = 0 and state that ‘(x – a) is a factor’. 2 5(b) Use f ( 3 ) = 0 M1 2  33 − 4  32 + 3 p + q = 0 Obtain 3 p + q = −18 A1 OE Powers should be evaluated but do not need to be simplified. Use f  ( 3 ) = 0 M1 54 − 24 + p = 0 Obtain p = −30 A1 OE Obtain p = −30, q = 72 A1 Correct only. Alternative Method for Question 5(b) 2 M1 Q  0 Attempt division of f(x) by( x − 3) as far as ( 2 x + Q ) Obtain quotient ( 2 x + 8 ) A1 Equate linear remainder to zero and compare coefficients M1 2 Or expand ( 2 x + 8 )( x − 3) and compare coefficients. Method to obtain an equation in p or q. obtain one of p = −30, q = 72 A1 Obtain p = −30, q = 72 A1 Correct answers only. 5(b) Alternative Method 2 for Question 5(b) Use f ( 3 ) = 0 M1 54 − 36 + 3 p + q = 0 Obtain 3 p + q = −18 A1 OE Powers evaluated. 2 M1 Using this as part of a hybrid method they need to be Divide f(x) by( x − 3) as far as ( 2 x + Q ) and form an equation in p only or q working towards a second equation by considering only or in p and q coefficients in the remainder or use f ( −4 ) = 0 or equivalent for their ( 2 x + 8 ) . Obtain correct equation in p or q A1 Obtain p = −30, q = 72 A1 Correct answer only. Alternative Method 3 for Question 5(b) f  ( x ) = 6 x 2 − 8 x + p B1 = ( x − 3 )( 6 x + 10 ) M1 Use the factor x − 3. Obtain p = −30 A1 2 3 2 M1 2 = 2 x − 4 x − 30 x + q Use the factor ( x − 3) . f ( x ) = ( x − 3) ( 2 x + Q ) ( )  Q = 8, q = 72 A1 5(b) Alternative Method 4 for Question 5(b) 2 M1 A = 2 can be found by inspection. Expand f ( x ) = ( x − 3) ( Ax + B ) and compare at least one coefficient other than for x3 Obtain q = 9 B and p = 9 A − 6 B A1 or obtain B = 8 Use their A and B to solve for p or q M1 A = 2 B = 8 Obtain one of p = −30, q = 72 A1 Obtain p = −30, q = 72 A1 Correct answer only. 5

More questions on Quadratics

6 (a) By sketching a suitable pair of graphs, show that the equation cot 2x = 2 sin 2x - 1 has exactly one root in the interval 0 1 x 1 1 r . [2] 2 (b) Show by calculation that the root is in the interval 0 .4 1 x 1 0 .6 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 -1 1 = (c) Use the iterative formula x tan e o to calculate the root correct to 2 decimal n + 1 2 2 sin 2 x - 1 n places. Give the result of each iteration to 4 decimal places. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Sketch a relevant graph, e.g. y = cot 2 x B1 2 cot(2x) Scales must be correct if seen. Do not need a full scale, but do need to be able to verify the key points mentioned below: 1 For y = cot2 x, correct intercept on x-axis, asymptotes at x = 0, x = 12 π implied 2sin(2x)-1 (condone large gap between the curve and the asymptote on the RHS). Must be continuous. O π π 4 2 For 2sin2 x − 1, correct shape and position. No incorrect curvature. Correct maximum implied. Intercept at (0, -1). 1 Sketch a second relevant graph, e.g. y = 2sin2 x − 1 and justify the given B1 Need to see enough of both graphs to confirm there is statement not a second intersection in the given interval. Need to mark the intersection with a dot or a cross, or say that the root lies at the point of intersection (or equivalent). 2 6(b) Calculate the values of a relevant expression or pair of expressions at M1 E.g. 0.9712  0.435and 0.389  0.864 x = 0.4and x = 0.6 0.537  0 and − 0.475  0 Or using the iterative formula, 0.18 > 0, –0.17 < 0 Allow working on a smaller interval contained within the given interval Or using tan 2 x ( 2sin 2 x − 1) −=1 0 – 0.55 < 0, and 1.22 > 0. M0 if working in degrees. Justify the given statement with correct calculated values A1 Values to 2 sf or better. Accept values with statement f ( 0.4 )  f ( 0.6 )  0. 2 6(c) 1 −1  1  M1 M0 if working in degrees. Use the iterative process xn +1 = tan   correctly at least once Need to get as far as a second iteration completed. 2  2sin2 xn − 1  Obtain final answer 0.49 A1 Show sufficient iterations to 4 d.p. to justify 0.49 to 2 d.p. or show that there A1 E.g. 0.5,0.4858,0.4967,0.4883,0.4947,.... is a sign change in ( 0.485, 0.495 ) or 0.4, 0.5804, 0.4378, 0.5395, 0.4595, 0.5188, 0.4726, 0.5075, 0.4804, 0.5009, 0.4851, 0.4972, 0.4879, –0.0157 < 0, 0.355 > 0 0.4950, 0.4895, 0.4937 or 0.6, 0.4291, 0.5483, 0.4544, 0.5235, 0.4695, 0.5101, 0.4786, 0.5025, 0.4840, 0.4981, 0.4872, 0.4955, 0.4891, 0.4940 The last 2 options are a lot of iterations. Allow the marks if they start correctly and finish correctly. Condone if there are some missing in the middle. 3

More questions on Trigonometry

7 The equation of a curve is 2y 3 - 3 x 2 y - x 3 = 16 . dy x 2 + 2 xy (a) Show that = . [4] dx 2y 2 - x 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the coordinates of the points on the curve at which the normal is parallel to the y-axis. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) State or imply 6 y 2 dy as the derivative of 2 y 3 B1 dx State or imply 3 x 2 dy + 6 xy as the derivative of 3x 2 y B1 dx Complete the differentiation and equate the derivative of the LHS to zero and M1 d y Needs to be clear how the value of was obtained, d y d x solve for (the = 0 can be implied) d x e.g. collecting like terms and using brackets. 2 A1 AG dy x + 2 xy Obtain = from correct working 2 dx 2 y 2 − x 2 dy 2 xy + x Accept = . dx 2 y 2 − x 2 No incorrect statements, e.g. ‘cancel by 3’ or ‘divide top and bottom by 3’ are acceptable, but to have dy 3 x 2 + 6 xy = and say “divide by 3” at the final step dx 6 y 2 − 3 x 2 scores A0. 4 7(b) Equate derivative to 0 and solve for x or for x in terms of y. *M1 E.g., x 2 + 2 xy = 0  x = 0 or x = −2 y. Must be using the numerator. Obtain ( 0, 2 ) B1 Allow if the values are stated separately. Do not ISW. Use their x = −2 y to form an equation in one unknown dM1 E.g., 2 y 3 − 12 y 3 + 8 y 3 = 16 or Or equivalent e.g. substitute y = −x2 2x  1 3   1 3  3 −2  x  + 3  x  − x = 16.  8   2  Obtain ( 4, − 2 ) A1 Allow if the values are stated separately. Do not ISW 4

More questions on Differentiation

8 (a) Prove the identity sin 4x / 4 sin x `2 cos 3 x - cos xj. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 4 r (b) Hence find the exact value of cos 3 x sin 4x dx. [5] y 0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) Use correct double angle formula to expand sin4x *M1 2sin2 x cos2 x OE OE. 2cos 2 x − 1 Use correct double angle formulae to obtain an expression in sin x and cos x dM1 E.g., 4sin x cos x ( ) 3 A1 AG Obtain sin4 x  4sin x 2cos x − cos x from fully correct working ( ) Alternative Method for Question 8(a) Use correct double angle formula to rearrange the RHS *M1 2 E.g. 2sin2 x 2cos x − 1 ( ) Use correct double angle formula to obtain an expression in sin2 x and cos2 x dM1 E.g. 2sin2 x cos2 x 3 A1 AG Obtain sin4 x  4sin x 2cos x − cos x from fully correct working ( ) Alternative Method 2 for Question 8(a) Use correct angle sum formulae to expand sin4x as far as an expression in *M1 E.g. sin x ( cos x cos2 x − sin x sin2 x ) sin x, cos x, sin2 x and cos2 x + cos x ( sin x cos2 x + cos x sin 2 x ) Use correct double angle formulae to obtain an expression in sin x and cos x dM1 2 3 E.g., sin x cos x 2cos x − 1 − 2sin x cos x ( ) + sin x cos x 2cos 2 x − 1 + 2sin x cos 3 x ( ) 3 A1 Obtain sin4 x  4sin x 2cos x − cos x from fully correct working ( ) 3 8(b) 6 4 B1 6 4 sin x cos x dx sin x cos x dx 8 Use the identity to obtain p  sin x cos x dx − 4   sin x cos x dx + q  Accept terms of the correct form but without the integral signs or the dx. Obtain r cos 7 x + s cos 5 x B1 Obtain − 8 cos 7 x + 4 cos 5 x B1 If using the substitution u = cos x, accept the form 7 5 8 7 4 5 − u + u . 7 5 Use the correct limits correctly in an expression of the form r cos 7 x + s cos 5 x M1 OE 8 1 4 1 8 4 Must be evaluated, e.g. −  +  + − . 7 8 2 5 4 2 7 5 1 A1 Or exact simplified equivalent (e.g. as 2 terms). Obtain 2 + 12 35 ( ) 5

More questions on Integration

x 2 + 4 ax + 6a 2 9 Let f ( x) = , where a is a positive constant. ( x + 2 a)( x + 3 a) (a) Express f ( x) in partial fractions. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ a (b) Hence find the exact value of f ( x) d x . Give your answer in the form a ( p + ln q) , where p and q -ya are rational. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 9(a) B C B1 State or imply the form A + + x + 2 a x + 3a Use a correct method for finding a coefficient M1 Obtain one of A = 1, B = 2a and C = −3a A1 SC: If B0 is scored because of an omission of ‘A’, then maximum M1A1 is available for one constant correct. SC: If they substitute a value for a, or if their working implies the use of a = 1, they can score maximum B1M1. Obtain a second value A1 Obtain the third value A1 Alternative Method for Question 9(a) linear expression B1 − ax State or imply 1 + 1 + ( x + 2 a )( x + 3a ) ( x + 2 a )( x + 3a ) B C B1 State or imply the form 1 + + x + 2a x + 3a Use a correct method for finding B or C M1 Obtain one of B = 2a and C = −3a A1 Obtain the second value A1 5 9(b) Integrate and obtain terms Ax + B ln ( x + 2 a ) + C ln ( x + 3a ) B2FT B1 for any two terms correct, B2 for all three terms correct. The FT is on A, B and C: x + 2 a ln ( x + 2 a ) − 3a ln ( x + 3a ) . Allow for FT on a split completed in (b). Substitute limits correctly in an integral containing at least 2 terms from the M1 3 4 E.g. 2 a + 2 a ln − 3a ln ( 1 ) ( 2 ) form rx + s ln ( x + 2 a ) + t ln ( x + 3a ) 9 A1 Accept equivalent fractions with integers. 2 + ln Obtain a from correct working ( ( 8 ) ) 9 2 + ln A0 XP if a is following an error in (a). ( 8 ( ) ) 4

More questions on Algebra